Problem: Find the sum of the infinite series $1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$.
Let
\[S = 1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\dotsb.\]Then
\[1998S = 1998 + 2 + \frac{3}{1998} + \frac{4}{1998^2} + \dotsb.\]Subtracting these equations, we get
\[1997S = 1998 + 1 + \frac{1}{1998} + \frac{1}{1988^2} + \dotsb = \frac{1998}{1 - 1/1998} = \frac{3992004}{1997},\]so $S = \boxed{\frac{3992004}{3988009}}.$